Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors]

or

out the rectangle B DOG. And now, by the called logarithms are just the lengths of the difhelp of this curve, we have an easy way of con ferent parts of this line measured on a scale of ceiving and computing the motion of a body equal parts. through the air. For the subtangent of our curve Reasons of convenience have given rise to now presents twice the height through which the another set of logarithms : these are suited to a ball must fall in vacuo, in order to acquire the logistic curve whose subtangent is only

19000

of terminal velocity; and therefore serves for a scale the ordinate tv, which is equal to the side of the on which to measure all the other representatives hyperbolic square, and which is assumed for the of the motion.

unit of number. We shall suit our applications It remains to make another observation on the of the preceding investigation to both these, and curve Lp K, which will save us all the trouble of shall first use the common logarithms whose subgeographical operations, and reduce the whole to tangent is 0.43429. The whole subject will be a very simple arithmetical computation. In con best illustrated by taking an example of the difstructing this curve we were limited to no par- ferent questions which may be proposed : Recolticular length of the line LR, which represented

u the space A CDB; and all that we had to take lect that the rectangle A COF is

- 2 a, or care of was, that when 0 C, Ox, Og, were taken

or E, for a ball of cast iron one inch diameter, in geometrical progression, Ms, M7, should be

ud in arithmetical progression. The abscissæ having and, if it has the diameter d, it is or 2 a d, ordinates equal to ps, ni, &c., might have been

or E d. twice as long as is shown in the dotted curve which is drawn through L. All the lines which I. It may be required to determine what will serve to measure the hyperbolic spaces would be the space described in a given time t by a then have been doubled. But NI would also ball setting out with a given velocity V, and have been doubled, and our proportions would what will be its velocity v at the end of that time. have still held good; because this sub-tangent Here we have NI: MÍ=ACOF:BDCA; is the scale of measurement of our figure, as E

now N I is the subtangent of the logistic curve; or 2 a is the scale of measurement for the M I is the difference between the logarithms of motions.

O D and OC; that is, the difference between Since then we have tables of logarithms calcu- the logarithms of e + tand e; ACOF is 2 ad,

u'd lated for every number, we may make use of

or Ed. Therefore by common logathem instead of this geometrical figure, which 8 still requires considerable trouble to suit it to rithms 0.43429 : log. e+t-- log.e = lad:S,= every case.

There are two sets of logarithmic space described, tables in common use. One is called a table of

ett hyperbolic or natural logarithms. It is suited

or 0.43429 : log. = 2 ad: S, to such a curve as is drawn in the figure, where

2 ad

ett the subtangent is equal to that ordinate 7 v which and S.

0.43429 corresponds to the side = 0) of the square rano inserted between the hyperbola and its assymp- by hyperbolic logarithms S= 2 ad x log.

ett totes. This square is the unit of surface, by which the hyperbolic areas are expressed; its Let the ball be a twelve pounder ; the initial side is the unit of length, by which the lines be- velicity 1600 feet, and the time twenty seconds. longing to the hyperbola are expressed; TV is=

2 ad. 1, or the unit of numbers to which the logarithms

We must first find e, which is

V. are suited, and then I N is also 1. Now the

Therefore, log. 2 a

+ 3:03236 square 0701 being unity, the area B ACD

log. d (4, 5)

+ 0.65321 will be some number; # () being also unity, OD

log. V. (1600)

- 3.20412 is some number : call it x. Then, by the nature of the hyperbola, O B:07=#0: DB; that

Log. of 3",03,=e

0.48145 is, * :1=1: so that D B is Now, And ett is 23”,v3, of which the log. is 1.36229

from which take the log. of e

0:48145 calling D di the area, B D d b, which is the fluxion (ultimately) of the hyperbolic area, is

remains the log of e+t

0.88084 Now in the curve LpK, M I has the same ratio

This must be considered as a common number to NT that BAC D has to 0107. Therefore,

2 ad if there be a scale of which N I is the unit, the by which we are to multiply

Therefore number on this scale corresponding to M I has add the logarithms of 2 ad

0.43429" the same ratio to which the number measuring

+ 3.68557

ett BAC D has to 1; and I i, which corresponds

log.

+9.94490 to BD db, is the Auxion (ultimately) of MI;

log. 0:43429

-9.63778 Therefore, if M I be called the logarithm of x',

Log. S. 9833 feet

3.99269 is properly represented by the fluxion of MI. In short, the line M I is divided precisely as the

For the final ve.ocity, line of numbers on a Gunter's scale, which is OD:0C = AC:BD, or e+t:e=V: v. therefore a line of logarithms; and the numbers 23",03 : 3",03= 1600 : 2102,=v.

e

[ocr errors]
[ocr errors]
[ocr errors]

1

is

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

e

e

T.

m

е

V.

V

e

The ball bas therefore gone 3278 yards, and

11:37 its velocity is reduced from 1600 to 2io.

0.75, and m is = 1.577.

7.21 The gradual progress of the ball, during some

Therefore log. 2 a 3.03236 seconds of its motion, is as follows

d 9.87506
S. Diff. V. Diff.

0.19782
397
1" 1383

1203
1073
239

Log. 2 adm 3.10524
2 2456

964
880
160

and 2 a d m = 1274.2. Now 1274:2 : 100 = 3 3336

804
744
114

ett:

2 adm 4080 690 0:43429 : 0·03408= log.

Bute = 645

86 5 4725

604
569

67
6 5294
537

= 0.763, and its logarithm 9.88252, which,

added to 0.03408, gives 9.91660, which is the The first column shows the time of the motion; log, of e+t, = 0·825, from which take e, and the second the space described ; the third the

62 differences of the spaces, showing the motion there remains t=0":062, or

of a second, during each successive seconds; the fourth the for the time of passage. Now, to find the re

1000 velocity at the end of the time t; and the last maining velocity, say 825 : 763=1670 : 1544, the differences of velocity, showing its diminution in each successive second. At the distance of maining velocity was only 1425, the ball having

But in Mr. Robins's experiment the re1000 yards the velocity is reduced to one-half, and lost 245 ; whereas by this computation it should at the distance of less tnan a mile it is reduced have lost only 126. It appears, therefore, that to one-third.

the resistance is double of what it would have II. Required to determine the distance at been if the resistance increased in the duplicate which the initial velocity V is reduced to any other quantity v. This question is solved in the proportion of the velocity. Mr. Robins says it

is nearly triple. But he supposes the resistance very same manner by substituting the logarithms of V and v for those of e+t and e; for AC: periment, so often mentioned, fully warrants.

to slow motions much smaller than his own ex

AC BD= OD : 0 C, and therefore log.

The time e in which the resistance of the air BD

would extinguish the velocity is 0":763. Gravity, OD

V
log. or log. =log.

ett
Required to

or the weight of the bullet, would have done it
1670

52

in determine the distance which in the velocity of

or 52'; therefore the resistance is 32

0.763 1780 of a twenty-four pound ball (which is the times, or nearly sixty-eight times its weight, by medium velocity of such a ball discharged with this theory, or 5.97 pounds. If we calculate sixteen pounds of powder), will be reduced to from Mr. Robins's experiment, we must say log. 1500. Here d is 5.68, and therefore

V the logarithm of 2 ad is

+ 3.78671 -: 0:43429= 100 : e V, which will be 630·23, Log. = 0·07433, of which the log. is +8.87116

630.23

52
and er
= 0":3774, and

gives 1670

-9.63778 Log. 0.43429

0:3774

138 for the proportion of the resistance to the Log. 1047-3, feet or 349 yards

weight, and makes the resistance 12:07 pounds, 3.02009

fully double of the other. This reduction will be produced in about seven

With this velocity, which greatly exceeds eighths of a second.

that with which the air can rush into a void, III. To determine the time which a ball, be- there must be a statical pressure of the atmoginning to move with a certain velocity, employs sphere equal to six pounds and a half. This will in passing over a given space, and the diminu- make up the difference; and allows us to contion of velocity which it sustains from the resist- clude that the resistance, arising solely from the auce of the air; proceed thus :

motion communicated to the air, follows very ett 2ad:S=0.43429 : log =t. Then to log. nearly the duplicate proportion of the velocity.

The next experiment, with a velocity of 1690 ett add log. e, and we obtain log. e+t, and feet, gives a resistance equal to 157 times the

weight of the bullet, and this bears a much ett; from which if we take e we have t. Then, greater proportion to the former than 1690 does to find v, say ettie= =V:v.

to 16702 ; which shows that, although these exThese examples may be concluded by apply- periments clearly demonstrate a prodigious auging this last rule to Mr. Robins's experiments mentation of resistance, yet they are by no on a musket bullet of three-fourths of an inch means susceptible of the precision which is in diameter, which had its velocity reduced from necessary for discovering the law of this aug1670 to 1425 by passing through 100 feet of air. mentation, or for a good foundation of practical This we do to discover the resistance which it rules; and it is still greatly to be wished that a sustained, and compare it with the resistance to more accurate mode of investigation could be a velocity of one foot per second. We must discovered. first ascertain the first term of our analogy. The We have thus explained, in detail, the princiball was of lead, and therefore 2 a must be mul- ples and the process of calculation for the simple tiplied by d and by m, which expresses the ratio case of the motion of projectiles through the air. of the density of lead to that of cast iron, d is The learned reader will think that we have been

[ocr errors]

V

e

e

[ocr errors]

.

[ocr errors]

that is,

[ocr errors]

unreasonably prolix, and that the whole might the hyperbolic logarithm of the quantity annexea have been comprised in less room, by taking the to it, and I may be used as to express its comalgebraic method. We acknowledge that it might mon logarithm. See article Fluxions. have been done even in a few lines. But we have . The constant quantity C for completing the observed, and our observation has been confirmed Auent is determined from this consideration, that by persons well versed in such subjects, that in the space described is o, when the velocity is o : all cases where the fluxionary process introduces

u the fluxion of a logarithm, there is a great want C. x Lu= 0, and C =

xLvu, of distinct ideas to accompany the hand and eye.

8

8 The solution comes out by a sort of magic or and the complete fluent S legerdemain, we cannot tell either how or why.

8 We therefore thought it necessary to furnish the reader with distinct conceptions of the things LviLvP_v=

=-x and quantities treated of. For this reason, after

8 showing, in Sir Isaac Newton's manner, how the

u?

uo Χλν

or (putting M for spaces described in the retarded motion of a pro- 0.43429 g,

u?-09 jectile followed the proportion of the hyperbolic 0:43429, the modulus or subtangent of the comareas, we showed the nature of another curve, where lines could be found which increase in the mon logistic curve) Χλ.

Mg

u-u very same manner as the path of the projectile increases; so that a point describing the abscissa This equation establishes the relation between MI of this curve moves precisely as the projec- the space fallen through and the velocity acquired tile does. Then, discovering that this line is the

u? same with the line of logarithms on a Gunter's by the fall. We obtain by it scale, we showed how the logarithm of a number really represents the path or space described by and 298

=L or, which is still more the projectile.

*ua_01

M X 2 g S Sect. V.-OF THE PERPENDICULAR ASCENTS convenient for us,

uo

u_02 AND DESCENTS OF HEAVY PROJECTILES.

equal to the logarithm of a certain number : Having thus enabled the reader to conceive therefore having found the natural number cordistinctly the quantities employed, we shall leave

M X 2g S the geometrical method, and prosecute the rest responding to the fraction

consider

u? of the subject in a more compendious manner. it as a logarithm, and take out the number corWe are next to consider the perpendicular ascents and descents of heavy projectiles, where responding to it: call this n. Then, since n is the resistance of the air is combined with the ac- equal to we have nu’-nv=u?, and nution of gravity: and we shall begin with the

u202' descents.

wox11-1 Let u, as before, be the terminal velocity, and u'en v?, or n v?=u'Xn-v, and v? = g the accelerating power of gravity : when the To expedite all the computations on this subbody moves with the velocity u, the resistance is ject, it will be convenient to have multipliers equal to g; and in every other velocity v, we ready computed for M X 2 g, and its half, must have u? : v2 = 8:

1.44396 viz. 27794, whose log.is = r, for the resistand 13.897

1.14293 ance to that velocity. In the descent the body But v may be found much more expeditiously by is urged by gravity g, and opposed by the resist

observing that

is the secant of an : therefore the remaining accelerating

u_02

arch of a circle whose radius is a, and whose force, which we shall call f, is g

sine is v, or whose radius is unity and sine = g u-g v2 8 (u2-02) > or =f.

therefore, considering the above fraction as a lo

garithmic secant, look for it in the tables, and The fundamental theorem for varied motion is then take the sine of the arc of which this is the

v

vu fs=ui, and ś=

secant, and multiply it by u; the product is the f

velocity required.

υ An example may be given of a ball whose ter+ C. Now the fluent of 6 u2v

minal velocity is 6891 feet, and ascertain its ve

u?_02 is = -hyper. log. of vu?—v. For the fluxion locity after a fall of 1848 feet. Here, -Vi

u = 475200, and its log. = 5.67688 of vu—22 is and this divided by u = 689}

2.83844 vu-029

1.50515 the quantity Nuv>, of which it is the fluxion,

S 1848

3.26676 Üv gives precisely which is therefore the flux

Then log. 27:794

+1:44396

+ 3.26670 ion of its hyperbolical logarithm. Therefore

-5.67688 S= xLvu?--02 + C. Where L means 6

9.03378 Log. of 0-10809 = lo in

& v2

.

ance

2

V

[ocr errors]

or

u 3

[ocr errors]

U

Х

and s=

[ocr errors]
[ocr errors]

xf

.

[ocr errors][merged small][ocr errors][merged small]

n

[ocr errors]
[ocr errors]

u

x 8

LV

U-V

Mg

U-V=

[ocr errors]

therefore avuto

U-V

u

[ocr errors]

u

[ocr errors]

.

0-10809 is the logarithm of 1.2826 = n, and n and the space described, or the velocity acquired,

*X -1 1= 0.2826, and

must now be ascertained.

For this purpose we = 323.6°, = d; may use the other fundamental proposition of and v= 323.6.

varied motions fi=v, which, in the present In like manner, 0.054045 (which is half of 0-10809) will be found to be the logarithmic case, becomes i = ; therefore i= secant of 28°, whose sine 0.46947

multiplied by 689f gives 324 for the velocity. The process of Х

and t =

X fi this solution suggests a very perspicuous manner of conceiving the law of descent; and it may be

u+u

utv thus expressed : M is to the logarithm of the

-x Lu

Χλ ν

8 secant of an arch whose sine is ñ and radius 1 plete it, or rather C=0; for t must be=o when

This fluent needs no constant quantity to comas 2 a is to the height through which the body v=o. This will evidently be the case ; for then must fall to acquire the velocity v. Thus, to take the same example :

LV . 1. Let the height h be sought which will pro

This rule may be illustrated by the same exduce the velocity 323-62, the terminal velocity of ample. In what time will the body acquire the the ball being 689-34. Here 2 a, oris 14850, velocity 323.62 ? Here utv=1012.

8 323.62

365•72;

= 0.22122, and a and = 0·46947, which is the sine of 28°.

689.35 The logarithmic secant of this arch is 0.05407. (in feet and seconds) is 21" .542. Now, for Now M or 0-43429 : 0-05407 = 14850 : 1848, greater perspicuity, convert the equation i = the height wanted.

utv

XXV into a proportion : thus M: 2. Required the velocity acquired by the body Mg by falling 1848 feet. Say 14850 1848 = utu 0-43429 : 0·05407. Look for this number among

: t, and we have 0.43429 :

8 the logarithmic secants. It will be found at 28°, 0-22122 = 21" 542 : 10":973, the time required. of which the logarithmic sine is 9.67161

We should remember that the numbers or symAdd to this the log. of u

2.83844 bols which we call logarithms are really parts of

the line M I in the figure of the logistic curve, and The sum

2-51005 that the motion of a point in this line is precisely is the logarithm of 323-62, the velocity required. similar to that of the body. The marquis Poleni,

From these solutions we see that the acquired in a dissertation published at Padua in 1725, velocity continually approaches to, but never has with great ingenuity constructed logarithmics equals, the terminal velocity. For it is always suited to all the cases which can occur. expressed by the sine of an arch of which the terminal velocity is the radius.

It is easy to see that ✓utu is the cotangent The motion of a body projected downwards Dext merits consideration. While the velocity of the } complement of an arch whose radius is of projection is less than the terminal velocity, 1, and whose sine is -: For let KC (see diathe motion is determined by what we have already said ; for we must compute the height gram) be := u, and B E= v; then KD=utu, necessary for acquiring this velocity in the air,

H and suppose the motion to have begun there. But, if the velocity of projection be greater, this

B method fails. We pass it over (though not in the least more difficult than what has gone before) because it is of mere curiosity, and never occurs in any interesting case. We may just observe that, since the motion is swifter than the

K

D A terminal velocity, the resistance must be greater and DA=440. Join K B and B A, and draw than the weight, and the motion will be retarded.

CG parallel to K B. Now G A is the tangent The very same process will give us for the space of ; BA, =complement of H B. Then, by described S=

V being the similarity of triangles, GA: AC=AB:Bk, v?

AC velocity of projection greater than u. Now as =VAD:VDK = Numvivuto and

GA this space evidently increases continually (because

uto the brdy always falls, but does not become in

(= cotan. BA)=V ; therefore look

V?__U finite in any finite time), the fraction

v2_-u?

the natural sines, or for log. -among does not become infinite; that is, va does not become equal to u? : therefore, although the velo- the logarithmic sines, and take the logarithmic city V is continually diminished, it never be- cotangent of the half complement of the corcomes so small as u. Therefore u is a limit of responding arch. This, considered as a common diminution as well as of augmentation.

number, will be the second term of our proporThe relation between the time of the descerit tion. This is a shorter process than the foriner,

u

[ocr errors]
[ocr errors]
[ocr errors]
[blocks in formation]

u

2 Mgs

2Mgs

; we

[ocr errors]

in vacuo, as ú LV uto

u

u

[ocr errors]

LV uto plied by

Mg

LV

-Lv

LV uto

[ocr errors]
[ocr errors]

u'+23. )

By reversing this proportion we get the velocity

+Va

Therefore let u be the corresponding to a given time. To compare u2 +02 this descent of 1848 feet in the air with the fall of the body in vacuo during the same time, say number whose common logarithm is 21" •5422 : 10" •973 = 1848 : 1926.6, which

u?+ V2

uo+V makes a difference of seventy-nine feet.

shall have n=

and o

n Cor. 1. The time in which the body acquires and thus we obtain the relation of s and v, as in the velocity u by falling through the air is, to the the case of descents; but we obtain it still easier time of acquiring the same velocity by falling by observing that vu?+ V? is the secant of an

arch whose radius is u, and whose tangent is V, to v; for it would

and that vu tv2 is the secant of another arch

of the same circle whose tangent is v. acquire this velocity in racuo during the time,

Let the same ball be projected upwards with

the velocity 411.05 feet per second. Required and it acquires it in the air in the time L

V 8 the whole height to which it will rise ? Here 11+v

will be found the tangent of 30:48, the logarith

mic secant of which is 0.06606. This, multi2. The velocity which the body acquires by

u? utv

gives 2259 feet for the height. It falling through the air in the time

4-0 would have risen 2640 feet in a void. is, to the velocity which it would acquire in vacuo Suppose this body to fall down again. We

utv can compare the velocity of projection with the during the same time as v to u

i velocity with which it again reaches the ground. for the velocity would acquire in vacuo during

The ascent and descent are equal ; therefore

u? + V? Putu the time

which multiplies the constant factor must be u 8

the multi(because in any time

in the ascent, is equal to the velocity w is ac

u*_22 8

plier in the descent. The first is the secant of quired). Next, let a body whose terminal velocity is u

an arch whose tangent is V; the other is the se

cant of an arch whose sine is v. These secants be projected perpendicularly upwards, with any velocity V. It is required to determine the height the velocity of projection is to the final returning

are equal, or the arches are the same; therefore to which it ascends, so as to have any remaining velocity v, and the time of its ascent; as also the velocity as the tangent to the sine, or as the height and time in which its whole 'motion will radius to the cosine of the arch. Thus suppose

the body projected with the terminal velocity, or be extinguished. We have now

g (u? + v2)

for u

V=u; then v== If V=689, v=487. the expression of f; for both gravity and resistance act now in the same direction and retard the Lastly, the relation of the space and the time g(uX v2) must be ascertained. Here

& (u++va)

is motion of the ascending body; therefore

u?

š

and s =

х =-vi, and s

Х

and

Х u’tva

12 + 2?

8

u? +22? uo v ☺ u?

)
+C,
x Lvu’t 23 +

Х

+C. Now (art. Fluxions) 8

8 C (see art. Fluxions). This must be = o at

is an arch whose tangent

and the beginning of the motion, that is, when v=V,

u? that is,

-* L Vu+ Va+C=0, or C= radius 1; therefore t= -x LVu+V, and the complete fluent will

+ C. This inust be = 0, when v = V, or 8 u2

V be s =

С
x arc. tan. -=0, and C

X arc. 8

V u? + V2

and the complete fluent is i= Х Χλν

8 8

Mg
u? +22

V

arc. tan. Let h be the greatest height to which the body will rise. Then s=h when v=0; and h =

u' within the brackets express a portion of the arch

of a circle whose radius is unity; and are thereu+12

u?
Χλν

fore abstract numbers, multiplying
x
We

which we Mg

have shown to be the number of upits of time in ;

therefore which a heavy body falls in vacuo from the u? tu?

height a, or in which it acquires the velocity u.

[ocr errors]

V2

u

[ocr errors]

u

[ocr errors][merged small][merged small][ocr errors]
[merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][ocr errors]
« ΠροηγούμενηΣυνέχεια »